31st Hellenic Mathematical Olympiad

$AEBF$ is cyclic ($BE$ and $BF$ are tangents and: $A\hat{E}B=A\hat{F}B=90^{\circ }$). Let $c_2$ be its circumcircle. The segment $CD$ is the common chord of the circles ($c$) and ($c_1$) and the segment $AB$ is the common chord of the circles ($c$) and ($c_2$). Finally, $EF$ is the common chord of the circles ($c_1$) and ($c_2$). Hence the chords $CD$, $AB$, $EF$ pass through the radical center, say $L$, of the three circles.

Since $BE$ and $BF$ are tangents of the circle $c_1(A,r)$, $AB$ is the bisector of the angle $E\hat{B}F$. Moreover, $AB$ is the bisector of the angle $C\hat{B}D$, because $A\hat{B}C$ and $A\hat{B}D$ are inscribed in the circle $c(O,R)$ and correspond to the equal arches $\widehat{AC}$ and $\widehat{AD}$. Hence $E\hat{B}C=F\hat{B}D$.

Figure 2

Using equality $E\hat{B}C=F\hat{B}D$, we will prove that the triangles $BCE$ and $BFD$ are similar. For that it is enough to prove that: $\frac{BC}{BE}=\frac{BF}{BD}\iff \frac{a}{x}=\frac{y}{a}\iff xy=a^2$, where $BE=BF=a$, $BC=x$ and $BD=y$. Similarly from $LCB\sim LAD$: $$\frac{LC}{LA}=\frac{CB}{AD}\Rightarrow \frac{LC}{LA}=\frac{x}{r}(1),$$ and from $LCA\sim LBD$, we have: $$\frac{LB}{LC}=\frac{BD}{CA}\Rightarrow \frac{LB}{LC}=\frac{y}{r}(2).$$

From (1) and (2) we have: $\frac{LB}{LA}=\frac{xy}{r^2}$ (A) and from the orthogonal triangle $ABE$ with $AL$ the altitude to the hypotenuse we get $\frac{LB}{LA}=\frac{E{B}^2}{E{A}^2}=\frac{a^2}{r^2}$. (B) From (A) and (B) we obtain: $xy=a^2$. Therefore $BCE\approx BFD$. From the equalities of their corresponding angles we find: $\hat{C}=\hat{F}$. Hence the quadrilateral $BCFM$ is cyclic.

Alternative solution: Let $Q=EF\cap CD$. From Pascal's theorem to degenerate hexagon EEDFFC, if $T=ED\cap CF$, we get that $T,B,M$ are collinear. Moreover, in the cyclic quadrilateral ECFD, the points $T,M$ are the common points of its opposite sides and $Q$ is the common point of its diagonals. Therefore the line TM is the polar of $Q$, and so $AQ\perp TM$ (A is the center of the circle) which means that $\hat{A}BT=90^{\circ }$. Since $AB\perp EF$, we get that $EF\parallel TM$. Hence we have: $$B\hat{M}C=M\hat{E}F=C\hat{F}B.$$

Figure 3